For any function \(\displaystyle f(x)=Ax^{n}\), the first derivative  \(\displaystyle f'(x)=nAx^{n-1}\).

Therefore, taking each term of \(\displaystyle f(x)\):

\(\displaystyle f(x)=\frac{x^{2}}{2}-x\)

\(\displaystyle f'(x)=\frac{2x^{2-1}}{2}-1(x^{1-1})\)

\(\displaystyle f'(x)=x-1\)

First apply the power rule to the entire function, multiplying it by its exponent and then subtracting 1 from the new exponent. Then apply the chain rule, multiplying the expression obtained from the power rule by the derivative of just the term inside of the parentheses. Remember the derivative of cos(x) is -sin(x):

\(\displaystyle f(x)=3(2x^3+\cos (x))^{50}\)

\(\displaystyle f'(x)=150(2x^3+\cos (x))^{49}\frac{d}{dx}(2x^3+\cos(x))\)

\(\displaystyle f'(x)=150(2x^3+\cos (x))^{49}(6x^2-\sin(x))\)

To find the derivative of this function, we simply need to use the Power Rule.  The Power rule states that for each term, we simply multiply the coefficient by the power to find the new coefficient.  We then decrease the power by one to obtain the degree of the new term.  

For example, with our first term, \(\displaystyle -3x^4\), we would multiply the coefficient \(\displaystyle (-3)\) by the power \(\displaystyle (4)\) to obtain the new coefficient of \(\displaystyle -12\).  We then decrease the power by one from 4 to 3 for the new degree.  Therefore, our new term is \(\displaystyle -12x^3\).  We then simply repeat the process with the remaining terms.

Note that with the second to last term, our degree is 1.  Therefore, multiplying the coefficient by the power gives us the same coefficient of 8.  When the degree decreases by one, we have a degree of 0, which simply becomes 1, making the entire term simply 8.

\(\displaystyle 8x^0=8(1)=8\)

With our final term, we technically have 

\(\displaystyle 17=17x^0\)

Therefore, multiplying our coefficient by our power of 0 makes the whole term 0 and thus negligible.

Our final derivative then is 

\(\displaystyle f'(x)=-12x^3+18x^2-2x+8\)

To find the first derivative, we will use the definition of the derivative which is,

\(\displaystyle f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}\).

In our case our,

\(\displaystyle f(x)=3x^2+4\) and

\(\displaystyle f(x+h)=3(x+h)^2+4=3(x^2+2xh+h^2)+4\).

Plugging these functions into our formula we get the following.

\(\displaystyle f'(x)=\lim_{h\rightarrow 0}\frac{3(x^2+2xh+h^2)+4-3x^2-4}{h}\)

\(\displaystyle f'(x)=\lim_{h\rightarrow 0}\frac{3x^2+6xh+3h^2+4-3x^2-4}{h}\)

\(\displaystyle f'(x)=\lim_{h\rightarrow 0}\frac{6xh+3h^2}{h}\)

\(\displaystyle f'(x)=\lim_{h\rightarrow 0}\frac{h}{h}\frac{6x+3h}{1}\)

\(\displaystyle f'(x)=\lim_{h\rightarrow 0}{6x+3h}\)

Now plugging in zero for h we are able to find the derivative of our function.

\(\displaystyle f'(x)=\lim_{h\rightarrow 0}{6x+3(0)}=6x\)

\(\displaystyle f'(x)=6x\)

In order to find the derivative we will need to use the power rule on each term. The power rule states,

\(\displaystyle f(x)=x^n \rightarrow f'(x)=nx^{n-1}\).

Applying this rule we get the following.

\(\displaystyle y=4x^5+3x^3+6x^2+1\)

\(\displaystyle y'=(4*5)x^{5-1}+(3*3)x^{3-1}+(6*2)x^{2-1}+(1)'\)

\(\displaystyle y'=20x^4+9x^2+12x+0\)

\(\displaystyle y'=20x^4+9x^2+12x\)

Because the original function is the quotient of 2 separate functions, we can use the Quotient Rule. 

Quotient Rule:

If

\(\displaystyle f(x)=\frac{u(x)}{v(x)}\),

then 

\(\displaystyle f'(x)=\frac{u'(x)v(x)-u(x)v'(x)}{(v(x))^2}\).

\(\displaystyle f(x)=\frac{5x^2}{3x+2}\) gives us 

\(\displaystyle u(x)=5x^2\), and \(\displaystyle v(x)=3x+2\).

Now

\(\displaystyle f'(x)=\frac{(5x^2)'(3x+2)-(5x^2)(3x+2)'}{(3x+2)^2}\)

Or 

\(\displaystyle f'(x)=\frac{10x(3x+2)-(5x^2)(3)}{(3x+2)^2}\).

Simplifying yields 

\(\displaystyle f'(x)=\frac{30x^2+20x-15x^2}{(3x+2)^2}\) or 

\(\displaystyle f'(x)=\frac{15x^2+20x}{(3x+2)^2}\).

We can factor out \(\displaystyle 5x\) to see that finally 

\(\displaystyle f'(x)=\frac{5x(3x+4)}{(3x+2)^2}\).

Because the two terms \(\displaystyle 3x+4\) and \(\displaystyle 3x+2\) are different, no more terms can cancel.

Using the Product Rule, which states,

\(\displaystyle [f(x)g(x)]'=f(x)g'(x)+g(x)f'(x)\)

we see that 

\(\displaystyle y'=(3x^2)'(\ln(3x))+(3x^2)(\ln(3x))'\)

\(\displaystyle y'=(6x)(\ln(3x))+(3x^2)(\frac{1}{3x}*3)\)

\(\displaystyle y'=6x\ln(3x)+3x\)

\(\displaystyle y'=3x(2\ln(3x)+1)\).

\(\displaystyle f(x)=(3x^2+4x+1)^5\) is a composition of 2 functions. The inner function is \(\displaystyle 3x^2+4x+1\), and the outer function is \(\displaystyle y^5\). We can use the Chain Rule.

The Chain Rule states,

\(\displaystyle [f(g(x))]'=f'(g(x))\cdot g'(x)\).

Therefore we get the following.

\(\displaystyle f'(x)=5*(3x^2+4x+1)^{5-1}(3x^2+4x+1)'\)

\(\displaystyle f'(x)=5(3x^2+4x+1)^4(6x+4)\)

The rest is simplifying...

\(\displaystyle f'(x)=5(3x^2+4x+1)^4*2(3x+2)\)

\(\displaystyle f'(x)=5*2(3x^2+4x+1)^4(3x+2)\)

\(\displaystyle f'(x)=10(3x^2+4x+1)^4(3x+2)\)

You can use the Chain Rule here, 

\(\displaystyle [f(g(x))]'=f'(g(x))\cdot g'(x)\)

\(\displaystyle y=(4\ln(2x))^3\)

In our case we get,

\(\displaystyle y'=3(4ln(2x))^2\cdot \left(4\frac{1}{2x}\cdot 2\right)\).

Remember the derivative of natural log is 

\(\displaystyle ln(u)\rightarrow \frac{1}{u}\cdot \frac{du}{dx}\).

Simplifying further we get our final answer.

\(\displaystyle y'=\frac{12(4ln(2x))^2}{x}\)

To find the derivative of a polynomial function, remember to use the power rule. Take the derivative of each term individually, making the power the new coefficient and then decreasing the power by one. Let us show how this works for each term. 

\(\displaystyle x^2\rightarrow 2x^{(2-1)} \rightarrow 2x\)

\(\displaystyle 5x \rightarrow5x^{1-1} = 5x^0 = 5\)

Since \(\displaystyle 2\) is a constant, we know that the derivative of a constant is just \(\displaystyle 0\). 

Thus, our entire expression is \(\displaystyle 2x + 5\).